- Elimination reactions occur by either an E1 or E2 mechanism. E1 is a one-step reaction involving a carbocation intermediate, while E2 is a concerted, single-step reaction. - The E1 mechanism is favored by good leaving groups, stable carbocations, and weak bases. It is non-stereospecific and does not occur with primary alkyl halides. The E2 mechanism is favored by strong bases and polar aprotic solvents. It is stereospecific and proceeds through an anti-periplanar transition state. - Key factors that determine the mechanism include the stability of carbocation intermediates, the strength of the leaving group and base, and steric

2. - Elimination reaction is a type of organic reaction in which
two substituents are removed from a molecule in either
a one or two-step mechanism.
- The one-step mechanism is known as the E2 reaction,
and the two-step mechanism is known as the E1 reaction.
- In most organic elimination reactions, at least one
hydrogen is lost to form the double bond: the unsaturation
of the molecule increases. It is also possible that a molecule
undergoes reductive elimination, by which the valence of an
atom in the molecule decreases by two, though this is more
common in inorganic chemistry.

3. There are three fundamental events in these elimination reactions:
(i) Removal of a proton
(ii) formation of the CC π bond
(iii) breaking of the bond to the leaving group

4. E1 mechanism

5. Reaction influence the reaction pathway:
E1 mechanistic pathway is most common with:
o Good leaving groups
o Stable carbocations
o Weak bases.
1. E1 Reactions is Non-stereospecific- follows
Zaitsev (Saytseff) Rule
2. Does NOT occur with primary alkyl halides
(leaving groups)

7. 1. In Elimination Reactions, The “More Substituted”
Alkene Tends To Be The Major Product

9. 2. The Stability Of Alkenes Increases As C-H Bonds Are
Replaced With C–C Bonds

10. 3. The “Zaitsev Rule”: Elimination Will Occur Such
That The Hydrogen Is Removed From The Beta-
Carbon With The Fewest Hydrogens

11. (i) Effect of R-
1. Reactivity order: (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3-
2. In an E1 reaction, the rate determining step is the loss of the
leaving group to form the intermediate carbocation.
3. The more stable the carbocation is, the easier it is to form, and
the faster the E1 reaction.
4. The rate of an E1 reaction increases as the number of R groups
on the carbon with the leaving group increases.

13. Energy Profile for an E1 Reaction

14. (ii) Leaving Group (LG)
- The only event in the rate determining step of the E1 is breaking
the C-LG bond. Therefore, there is a very strong dependence on the
nature of the leaving group, the better the leaving group, the faster
the E1 reaction will be.
- In the acid catalysed reactions of alcohols, the -OH is protonated
first to give an oxonium ion, providing the much
better leaving group, a water molecule.
(iii) Base (B)
- Since the base is not involved in the rate determining step, the
nature of the base is unimportant in an E1 reaction.
- Favored by weaker bases such as H2O and ROH.
(iv) Type of Solvent
- Favored by polar protic solvents, which can stabilize the ionic
intermediates.

15. E1 Mechanism for Alcohols
Step 1:
An acid/base reaction. Protonation of the alcoholic oxygen to
make a better leaving group. This step is very fast and reversible.
The lone pairs on the oxygen make it a Lewis base.

16. Step 2:
Cleavage of the C-O bond allows the loss of the
good leaving group, a neutral water molecule, to
give a carbocation intermediate. This is the rate
determining step (bond breaking is endothermic)
Step 3:
An acid/base reaction. Deprotonation by a base (a
water molecule) from a C atom adjacent to the
carbocation center leads to the creation of the C=C

17. E1 Mechanism for Alkyl Halides
Step 1:
Cleavage of the polarised C-X bond allows
the loss of the good leaving group, a halide
ion, to give a carbocation intermediate. This is
the rate determining step (bond breaking is
endothermic)
Step 2:
An acid/base reaction. Deprotonation by a
base (here an alkoxide ion) from a C atom
adjacent to the carbocation center leads to the
creation of the C=C

18. E2 mechanism
1. E2 indicates an elimination, bimolecular reaction, where rate = k
[B][R-LG].
2. This implies that the rate determining step involves an interaction
between these two species, the base B, and the organic substrate,
R-LG
Removal of the proton, H+, by the base, loss of the leaving
group, LG, and formation of the π-bond

19. Kinetics – Second order
Mechanism
- Single step
- Stereospecific (Anti-periplanar geometry preferred, Syn-
periplanar geometry possible)
- Concerted - all bonds form and break at same time.
- Bimolecular - rate depends on concentration of both base and
substrate
- Favoured by strong bases.
- Favored by polar aprotic solvents.
- Better leaving group leads to faster reaction rates.
Identity of R group – More substituted halides react faster Rate:
R3CX > R2CHX > RCH2X

20. Effects of R-
- Reactivity order: (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3-
- In an E2 reaction, the reaction transforms 2 sp3 C atoms into
sp2 C atoms. This moves the substituents further apart
decreasing any steric interactions.
- So more highly substituted systems undergo E2 eliminations
more rapidly. This is the same reactivity trend as seen
in E1 reactions.
- As the number of R groups on the carbon with the leaving group
increases, the rate of the E2 reaction increases.

22. (ii) Leaving Group (LG)
- The C-LG bond is broken during the rate determining step,
so the rate does depend on the nature of the leaving
group.
- However, if a leaving group is too good, then an E1 reaction
may result.
- Rate of reaction follows the order Rate of reaction follows
the order, R−I > R−Br > R−Cl > R−F
(iii) Base (B)
- Stronger bases favor the reaction. Since the base is involved
in the rate determining step, the nature of the base is
very important in an E2 reaction.
- More reactive bases will favor an E2 reaction

23. E2 Mechanism
Stereochemistry of the E2 Reaction
- The transition state of an E2 reaction consists of four atoms
from the substrate (one hydrogen atom, two carbon atoms, and
the leaving group, X) aligned in a plane. There are two ways for
the C—H and C—X bonds to be coplanar.

24. 1. E2 elimination occurs most often in the anti periplanar geometry. This
arrangement allows the molecule to react in the lower energy staggered
conformation, and allows the incoming base and leaving group to be
further away from each other.
2. The anti periplanar geometry also allows direct interaction between the
bonding electrons of C — H bond and the antibonding orbital of the C
—X bond.
3. Diastereomeric starting compounds yield diastereomeric products after
an E2 reaction.